不需要转换为中间pandas的字符串，你可以使用pyspark用户定义函数（udf）来unquote引用的字符串：

from urllib.parse import unquote

df.withColumn('campaign', F.udf(unquote, F.StringType())('campaign'))

如果campaign列中有null值，那么在取消引用字符串之前必须进行null检查：

f = lambda s: unquote(s) if s else s
df.withColumn('campaign',  F.udf(f, F.StringType())('campaign'))

+-----------------+-----------------+
|          user_id|         campaign|
+-----------------+-----------------+
| alskd9239as23093|MM+|+Consórcios+%|
|  lfifsf093039388|      Aquisição+||
|kasd877191kdsd999|      Aquisição+||
+-----------------+-----------------+

这些应该是工作：

Spark 3.5+

F.url_decode('campaign')

Spark 3.4+

F.expr("url_decode(campaign)")

然而，Spark说你的值“MM+%7C+Cons%C3%B3rcios+%”格式不正确：
[CANNOT_DECODE_URL]无法解码提供的URL：MM+%7C+Cons%C3%B3rcios+%.请检查您输入的网址是否正确，然后重试。
当我尝试不使用该行时，它有效：

from pyspark.sql import functions as F
df = spark.createDataFrame(
    [('lfifsf093039388', 'Aquisi%C3%A7%C3%A3o+%7C'),
     ('kasd877191kdsd999', 'Aquisi%C3%A7%C3%A3o+%7C')],
    ['user_id', 'campaign'])

df.withColumn('campaign', F.url_decode('campaign')).show()
# +-----------------+-----------+
# |          user_id|   campaign|
# +-----------------+-----------+
# |  lfifsf093039388|Aquisição ||
# |kasd877191kdsd999|Aquisição ||
# +-----------------+-----------+